先化简,再求值【(1/X^2+3X+2)+(1/x^2+5X+6)+1/X^2+7X+12】÷(1/X^2+9X+20)其中x=根号2

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/01 05:46:50
先化简,再求值【(1/X^2+3X+2)+(1/x^2+5X+6)+1/X^2+7X+12】÷(1/X^2+9X+20)其中x=根号2

先化简,再求值【(1/X^2+3X+2)+(1/x^2+5X+6)+1/X^2+7X+12】÷(1/X^2+9X+20)其中x=根号2
先化简,再求值【(1/X^2+3X+2)+(1/x^2+5X+6)+1/X^2+7X+12】÷(1/X^2+9X+20)其中x=根号2

先化简,再求值【(1/X^2+3X+2)+(1/x^2+5X+6)+1/X^2+7X+12】÷(1/X^2+9X+20)其中x=根号2
原式={1/((X+1)(X+2))+1/((X+2)(X+3))+1/((X+3)(X+4))}/{1/((X+4)(X+5))}
={1/(X+1)-1/(X+2)+1/(X+2)-1/(X+3)+1/(X+3)-1/(X+4)}/{1/((X+4)(X+5))}
={1/(X+1)-1/(X+4)}/{1/((X+4)(X+5))}
=3/((X+1)(X+4))/{1/((X+4)(X+5))}
=3(X+5)/(X+1)
=3(5+根号2)/(1+根号2)
=(12根号2)-9

你去问数学家去吧

x=根号2
【1/(X^2+3X+2)+ 1/(x^2+5X+6)+1/(X^2+7X+12)】÷1/(X^2+9X+20)
= 【1/{(x+1)(x+2)} + 1/{(x+2)(x+3) } + 1/{(X+3)(x+4)}】 × (x+4)(x+5)
= 【1/(x+1) -1/(x+2) + 1/(x+2) - 1/(x+3) +1/(X+3) -1/(x+4...

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x=根号2
【1/(X^2+3X+2)+ 1/(x^2+5X+6)+1/(X^2+7X+12)】÷1/(X^2+9X+20)
= 【1/{(x+1)(x+2)} + 1/{(x+2)(x+3) } + 1/{(X+3)(x+4)}】 × (x+4)(x+5)
= 【1/(x+1) -1/(x+2) + 1/(x+2) - 1/(x+3) +1/(X+3) -1/(x+4)}】× (x+4)(x+5)
= 【1/(x+1) -1/(x+4)}】× (x+4)(x+5)
= { (x+4)-(x+1)}/{(x+1) (x+4)} × (x+4)(x+5)
= 3 (x+5) /(x+1)
= 3(根号2 +5)/(根号2+1)
= 3(根号2+5)(根号2-1)/(2-1)
= 3(4根号2 - 3)

收起

这么复杂你自己可好意思让别人化啊

[1/(X^2+3X+2)+1/(x^2+5X+6)+1/(X^2+7X+12)]÷[1/(X^2+9X+20)]
=[1/(X+1)(X+2)+1/(x+2)(X+3)+1/(X+3)(X+4)]÷[1/(X+4)(X+5)]
=[1/(X+1)-1/(X+2)+1/(x+2)-1/(X+3)+1/(X+3)-1/(X+4)]÷[1/(X+4)(X+5)]
=[1/(X+...

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[1/(X^2+3X+2)+1/(x^2+5X+6)+1/(X^2+7X+12)]÷[1/(X^2+9X+20)]
=[1/(X+1)(X+2)+1/(x+2)(X+3)+1/(X+3)(X+4)]÷[1/(X+4)(X+5)]
=[1/(X+1)-1/(X+2)+1/(x+2)-1/(X+3)+1/(X+3)-1/(X+4)]÷[1/(X+4)(X+5)]
=[1/(X+1)-1/(X+4)]÷[1/(X+4)(X+5)]
=[(X+4-X-1)/(X+1)(X+4)]÷[1/(X+4)(X+5)]
=[3/(X+1)(X+4)]÷[1/(X+4)(X+5)]
=3/(X+1)(X+4)*(X+4)(X+5)
=3(X+5)/(X+1)
=[3(X+1)+2]/(X+1)
=3(X+1)/(X+1)+2/(X+1)
=3+2/(X+1)
=3+2/(√2+1)
=3+2(√2-1)/(√2+1)(√2-1)
=3+2(√2-1)
=3+2√2-2
=1+2√2

收起